EECE 370 Problems Solution Manual, Fall 2011

i Solutions Manual to accompany Chapman Electric Machinery Fundamentals Fifth Edition Stephen J. Chapman BAE Systems Australia ii Solutions Manual to accompany Electric Machinery Fundamentals, Fifth Edition Copyright  2012 McGraw-Hill, Inc. All rights reserved. Printed in the United States of America. No part of this book may be used or reproduced in any manner whatsoever without written permission, with the following exception: homework solutions may be copied for classroom use. iii TABLE OF CONTENTS Preface iv 1 Introduction to Machinery Principles 1 2 Transformers 23 3 AC Machine Fundamentals 73 4 Synchronous Generators 81 5 Synchronous Motors 123 6 Induction Motors 152 7 DC Machinery Fundamentals 202 8 DC Motors and Generators 214 9 Single-Phase and Special Purpose Motors 276 A Review of Three-Phase Circuits 287 B Coil Pitch and Distributed Windings 295 C Salient-Pole Theory of Synchronous Machines 302 S1 Introduction to Power Electronics 308 E Errata 348 iv PREFACE TO THE INSTRUCTOR This Instructor’s Manual is intended to accompany the fifth edition of Electric Machinery Fundamentals. To make this manual easier to use, it has been made self-contained. Both the original problem statement and the problem solution are given for each problem in the book. This structure should make it easier to copy pages from the manual for posting after problems have been assigned. Many of the problems in Chapters 2, 4, 5, and 8 require that a student read one or more values from a magnetization curve. The required curves are given within the textbook, but they are shown with relatively few vertical and horizontal lines so that they will not appear too cluttered. Electronic copies of the corresponding open-circuit characteristics, short-circuit characteristics, and magnetization curves as also supplied with the book. They are supplied in as ASCII text files. Students can use these files for electronic solutions to homework problems. The ASCII files can be read into MATLAB and used to interpolate points along the curve. Each curve is given in ASCII format with comments at the beginning. For example, the magnetization curve in Figure P8-1 is contained in file p81_mag.dat. Its contents are shown below: % This is the magnetization curve shown in Figure % P8-1. The first column is the field current in % amps, and the second column is the internal % generated voltage in volts at a speed of 1200 r/min. % To use this file in MATLAB, type "load p81_mag.dat". % The data will be loaded into an N x 2 array named % "p81_mag", with the first column containing If and % the second column containing the open-circuit voltage. % MATLAB function "interp1" can be used to recover % a value from this curve. 0 0 0.0132 6.67 0.03 13.33 0.033 16 0.067 31.30 0.1 45.46 0.133 60.26 0.167 75.06 0.2 89.74 0.233 104.4 0.267 118.86 0.3 132.86 0.333 146.46 0.367 159.78 0.4 172.18 0.433 183.98 0.467 195.04 v 0.5 205.18 0.533 214.52 0.567 223.06 0.6 231.2 0.633 238 0.667 244.14 0.7 249.74 0.733 255.08 0.767 259.2 0.8 263.74 0.833 267.6 0.867 270.8 0.9 273.6 0.933 276.14 0.966 278 1 279.74 1.033 281.48 1.067 282.94 1.1 284.28 1.133 285.48 1.167 286.54 1.2 287.3 1.233 287.86 1.267 288.36 1.3 288.82 1.333 289.2 1.367 289.375 1.4 289.567 1.433 289.689 1.466 289.811 1.5 289.950 To use this curve in a MATLAB program, the user would include the following statements in the program: % Get the magnetization curve. Note that this curve is % defined for a speed of 1200 r/min. load p81_mag.dat if_values = p81_mag(:,1); ea_values = p81_mag(:,2); n_0 = 1200; The solutions in this manual have been checked twice, but inevitably some errors will have slipped through. If you locate errors which you would like to see corrected, please feel free to contact me at the address shown below, or at my email address schapman@tpgi.com.au. I greatly appreciate your input! My physical and email addresses may change from time to time, but my contact details will always be available at the book’s Web site, which is http://www.mhhe.com/chapman/. Thank you. Stephen J. Chapman Melbourne, Australia March 31, 2011 Chapter 1: Introduction to Machinery Principles 1-1. A motor’s shaft is spinning at a speed of 1800 r/min. What is the shaft speed in radians per second? SOLUTION The speed in radians per second is   1 min 2 rad 1800 r/min 188.5 rad/s 60 s 1 r              1-2. A flywheel with a moment of inertia of 4 kg  m2 is initially at rest. If a torque of 6 N  m (counterclockwise) is suddenly applied to the flywheel, what will be the speed of the flywheel after 5 s? Express that speed in both radians per second and revolutions per minute. SOLUTION The speed in radians per second is:   2 6 N m 5 s 7.5 rad/s 4 kg m t t J                The speed in revolutions per minute is:   1 r 60 s 7.5 rad/s 71.6 r/min 2 rad 1 min n             1-3. A force of 10 N is applied to a cylinder, as shown in Figure P1-1. What are the magnitude and direction of the torque produced on the cylinder? What is the angular acceleration  of the cylinder? SOLUTION The magnitude and the direction of the torque on this cylinder is: , CCW sin ind    rF    ind 0.15 m 10 N sin 30 0.75 N m, CCW      The resulting angular acceleration is: 2 2 0.75 N m 0.188 rad/s 4 kg m J        1-4. A motor is supplying 50 N  m of torque to its load. If the motor’s shaft is turning at 1500 r/min, what is the mechanical power supplied to the load in watts? In horsepower? SOLUTION The mechanical power supplied to the load is 1    1 min 2 rad 50 N m 1500 r/min 7854 W 60 s 1 r P                  1 hp 7854 W 10.5 hp 746 W P         1-5. A ferromagnetic core is shown in Figure P1-2. The depth of the core is 5 cm. The other dimensions of the core are as shown in the figure. Find the value of the current that will produce a flux of 0.005 Wb. With this current, what is the flux density at the top of the core? What is the flux density at the right side of the core? Assume that the relative permeability of the core is 800. SOLUTION There are three regions in this core. The top and bottom form one region, the left side forms a second region, and the right side forms a third region. If we assume that the mean path length of the flux is in the center of each leg of the core, and if we ignore spreading at the corners of the core, then the path lengths are = 2(27.5 cm) = 55 cm, = 30 cm, and = 30 cm. The reluctances of these regions are: 1l 2l 3l      1 7 0.55 m 72.9 kA t/Wb 800 4 10 H/m 0.05 m 0.15 m r o l l A A            R      2 7 0.30 m 59.7 kA t/Wb 800 4 10 H/m 0.05 m 0.10 m r o l l A A            R      3 7 0.30 m 119.4 kA t/Wb 800 4 10 H/m 0.05 m 0.05 m r o l l A A            R The total reluctance is thus TOT 1 2 3 72.9 59.7 119.4 252 kA t/Wb         R R R R and the magnetomotive force required to produce a flux of 0.005 Wb is    0.005 Wb 252 kA t/Wb 1260 A t      F R and the required current is 1260 A t 2.5 A 500 t i N     F The flux density on the top of the core is 2    0.005 Wb 0.67 T 0.15 m 0.05 m B A     The flux density on the right side of the core is    0.005 Wb 2.0 T 0.05 m 0.05 m B A     1-6. A ferromagnetic core with a relative permeability of 1500 is shown in Figure P1-3. The dimensions are as shown in the diagram, and the depth of the core is 5 cm. The air gaps on the left and right sides of the core are 0.050 and 0.070 cm, respectively. Because of fringing effects, the effective area of the air gaps is 5 percent larger than their physical size. If there are 300 turns in the coil wrapped around the center leg of the core and if the current in the coil is 1.0 A, what is the flux in each of the left, center, and right legs of the core? What is the flux density in each air gap? SOLUTION This core can be divided up into five regions. Let be the reluctance of the left-hand portion of the core, be the reluctance of the left-hand air gap, R be the reluctance of the right-hand portion of the core, be the reluctance of the right-hand air gap, and be the reluctance of the center leg of the core. Then the total reluctance of the core is 1 R 3 2 R R4 5 R    1 2 3 4 TOT 5 1 2 3 4        R R R R R R R R R R      1 1 7 0 1 1.11 m 168 kA t/Wb 1500 4 10 H/m 0.07 m 0.05 m r l   A        R      2 2 7 0 2 0.0007 m 152 kA t/Wb 4 10 H/m 0.07 m 0.05 m 1.05 l  A        R      3 3 7 0 3 1.11 m 168 kA t/Wb 1500 4 10 H/m 0.07 m 0.05 m r l   A        R      4 4 7 0 4 0.0005 m 108 kA t/Wb 4 10 H/m 0.07 m 0.05 m 1.05 l  A        R 3      5 5 7 0 5 0.37 m 56.1 kA t/Wb 1500 4 10 H/m 0.07 m 0.05 m r l   A        R The total reluctance is       1 2 3 4 TOT 5 1 2 3 4 168 152 168 108 56.1 204 kA t/Wb 168 152 168 108                 R R R R R R R R R R The total flux in the core is equal to the flux in the center leg:    center TOT TOT 300 t 1.0 A 0.00147 Wb 204 kA t/Wb       F R The fluxes in the left and right legs can be found by the “flux divider rule”, which is analogous to the current divider rule.       3 4 left TOT 1 2 3 4 168 108 0.00147 Wb 0.00068 Wb 168 152 168 108              R R R R R R       1 2 right TOT 1 2 3 4 168 152 0.00147 Wb 0.00079 Wb 168 152 168 108              R R R R R R The flux density in the air gaps can be determined from the equation   BA :     left left eff 0.00068 Wb 0.185 T 0.07 cm 0.05 cm 1.05 B A         right right eff 0.00079 Wb 0.215 T 0.07 cm 0.05 cm 1.05 B A     1-7. A two-legged core is shown in Figure P1-4. The winding on the left leg of the core (N1) has 600 turns, and the winding on the right (N2) has 200 turns. The coils are wound in the directions shown in the figure. If the dimensions are as shown, then what flux would be produced by currents i1 = 0.5 A and i2 = 1.0 A? Assume r  = 1200 and constant. 4 SOLUTION The two coils on this core are would so that their magnetomotive forces are additive, so the total magnetomotive force on this core is       TOT 1 1 2 2 600 t 0.5 A 200 t 1.00 A 500 A t N i N i       F The total reluctance in the core is      TOT 7 0 2.60 m 76.6 kA t/Wb 1200 4 10 H/m 0.15 m 0.15 m r l   A        R and the flux in the core is: TOT TOT 500 A t 0.00653 Wb 76.6 kA t/Wb       F R 1-8. A core with three legs is shown in Figure P1-5. Its depth is 5 cm, and there are 100 turns on the leftmost leg. The relative permeability of the core can be assumed to be 2000 and constant. What flux exists in each of the three legs of the core? What is the flux density in each of the legs? Assume a 5% increase in the effective area of the air gap due to fringing effects. 5 SOLUTION This core can be divided up into four regions. Let be the reluctance of the left-hand portion of the core, be the reluctance of the center leg of the core, be the reluctance of the center air gap, and be the reluctance of the right-hand portion of the core. Then the total reluctance of the core is the reluctance of the left-hand leg plot the parallel combination of the reluctances of the right-hand and center legs: 1 R R2 3 R 4 R   2 3 TOT 1 2 3      R R R R R R R R 4 4      1 1 7 0 1 1.08 m 95.5 kA t/Wb 2000 4 10 H/m 0.09 m 0.05 m r l   A        R      2 2 7 0 2 0.34 m 18.0 kA t/Wb 2000 4 10 H/m 0.15 m 0.05 m r l   A        R      3 3 7 0 3 0.0005 m 51.0 kA t/Wb 4 10 H/m 0.15 m 0.05 m 1.04 l  A        R      4 4 7 0 4 1.08 m 95.5 kA t/Wb 2000 4 10 H/m 0.09 m 0.05 m r l   A        R The total reluctance is     2 3 4 TOT 1 2 3 4 18.0 51.0 95.5 95.5 135.5 kA t/Wb 18.0 51.0 95.0             R R R R R R R R The total flux in the core is equal to the flux in the left leg:    left TOT TOT 100 t 2.0 A 0.00148 Wb 135.5 kA t/Wb       F R The fluxes in the center and right legs can be found by the “flux divider rule”, which is analogous to the current divider rule.   4 center TOT 2 3 4 95.5 0.00148 Wb 0.00086 Wb 18.0 51.0 95.5          R R R R 6   2 3 right TOT 2 3 4 18.0 51.0 0.00235 Wb 0.00062 Wb 18.0 51.0 95.5            R R R R R The flux density in the legs can be determined from the equation   BA :    left left 0.00148 Wb 0.329 T 0.09 m 0.05 m B A        center center 0.00086 Wb 0.115 T 0.15 m 0.05 m B A        left right 0.00062 Wb 0.138 T 0.09 cm 0.05 cm B A     1-9. A wire is shown in Figure P1-6 which is carrying 2.0 A in the presence of a magnetic field. Calculate the magnitude and direction of the force induced on the wire. SOLUTION The force on this wire can be calculated from the equation       2 A 1 m 0.5 T 1.00 N, into the page i ilB      F l B 1-10. The wire is shown in Figure P1-7 is moving in the presence of a magnetic field. With the information given in the figure, determine the magnitude and direction of the induced voltage in the wire. SOLUTION The induced voltage on this wire can be calculated from the equation shown below. The voltage on the wire is positive downward because the vector quantity v B points downward. 7       ind cos 45 10 m/s 0.2 T 0.25 m cos 45 0.354 V, positive down e vBl         v B l 1-11. Repeat Problem 1-10 for the wire in Figure P1-8. SOLUTION The induced voltage on this wire can be calculated from the equation shown below. The total voltage is zero, because the vector quantity v B points into the page, while the wire runs in the plane of the page.       ind cos 90 1 m/s 0.5 T 0.5 m cos 90 0 V e vBl         v B l 1-12. The core shown in Figure P1-4 is made of a steel whose magnetization curve is shown in Figure P1-9. Repeat Problem 1-7, but this time do not assume a constant value of µr. How much flux is produced in the core by the currents specified? What is the relative permeability of this core under these conditions? Was the assumption in Problem 1-7 that the relative permeability was equal to 1200 a good assumption for these conditions? Is it a good assumption in general? 8 SOLUTION The magnetization curve for this core is shown below: 192 9 The two coils on this core are wound so that their magnetomotive forces are additive, so the total magnetomotive force on this core is       TOT 1 1 2 2 600 t 0.5 A 200 t 1.00 A 500 A t N i N i       F Therefore, the magnetizing intensity H is 500 A t 192 A t/m 2.60 m c H l      F From the magnetization curve, 0.17 T B  and the total flux in the core is     TOT 0.17 T 0.15 m 0.15 m 0.00383 Wb BA     The relative permeability of the core can be found from the reluctance as follows: A l r 0 TOT TOT     R  F Solving for µr yields         TOT -7 TOT 0 0.00383 Wb 2.6 m 704 500 A t 4 10 H/m 0.15 m 0.15 m r l A          F The assumption that r  = 1200 is not very good here. It is not very good in general. 1-13. A core with three legs is shown in Figure P1-10. Its depth is 5 cm, and there are 400 turns on the center leg. The remaining dimensions are shown in the figure. The core is composed of a steel having the magnetization curve shown in Figure 1-10c. Answer the following questions about this core: (a) What current is required to produce a flux density of 0.5 T in the central leg of the core? (b) What current is required to produce a flux density of 1.0 T in the central leg of the core? Is it twice the current in part (a)? (c) What are the reluctances of the central and right legs of the core under the conditions in part (a)? (d) What are the reluctances of the central and right legs of the core under the conditions in part (b)? (e) What conclusion can you make about reluctances in real magnetic cores? 10 SOLUTION The magnetization curve for this core is shown below: (a) A flux density of 0.5 T in the central core corresponds to a total flux of     TOT 0.5 T 0.05 m 0.05 m 0.00125 Wb BA     By symmetry, the flux in each of the two outer legs must be 1 2 0.000625 Wb    , and the flux density in the other legs must be    1 2 0.000625 Wb 0.25 T 0.05 m 0.05 m B  B   The magnetizing intensity H required to produce a flux density of 0.25 T can be found from Figure 1-10c. It is 50 A·t/m. Similarly, the magnetizing intensity H required to produce a flux density of 0.50 T is 75 A·t/m. The mean length of the center leg is 21 cm and the mean length of each outer leg is 63 dm, so the total MMF needed is TOT center center outer outer H l H l   F       TOT 75 A t/m 0.21 m 50 A t/m 0.63 m 47.3 A t      F  and the required current is TOT 47.3 A t 0.12 A 400 t i N     F (b) A flux density of 1.0 T in the central core corresponds to a total flux of     TOT 1.0 T 0.05 m 0.05 m 0.0025 Wb BA     By symmetry, the flux in each of the two outer legs must be 1 2 0.00125 Wb    , and the flux density in the other legs must be    1 2 0.00125 Wb 0.50 T 0.05 m 0.05 m B  B   11 The magnetizing intensity H required to produce a flux density of 0.50 T can be found from Figure 1-10c. It is 75 A·t/m. Similarly, the magnetizing intensity H required to produce a flux density of 1.00 T is about 160 A·t/m. Therefore, the total MMF needed is TOT center center outer outer H I H I   F       TOT 160 A t/m 0.21 m 75 A t/m 0.63 m 80.8 A t      F  and the required current is TOT 80.8 A t 0.202 A 400 t i N      This current is not twice the current in part (a). (c) The reluctance of the central leg of the core under the conditions of part (a) is:    TOT cent TOT 75 A t/m 0.21 m 12.6 kA t/Wb 0.00125 Wb       F R The reluctance of the right leg of the core under the conditions of part (a) is:    TOT right TOT 50 A t/m 0.63 m 50.4 kA t/Wb 0.000625 Wb       F R (d) The reluctance of the central leg of the core under the conditions of part (b) is:    TOT cent TOT 160 A t/m 0.21 m 13.4 kA t/Wb 0.0025 Wb       F R The reluctance of the right leg of the core under the conditions of part (b) is:    TOT right TOT 75 A t/m 0.63 m 37.8 kA t/Wb 0.00125 Wb       F R (e) The reluctances in real magnetic cores are not constant. 1-14. A two-legged magnetic core with an air gap is shown in Figure P1-11. The depth of the core is 5 cm, the length of the air gap in the core is 0.05 cm, and the number of turns on the coil is 1000. The magnetization curve of the core material is shown in Figure P1-9. Assume a 5 percent increase in effective air-gap area to account for fringing. How much current is required to produce an air-gap flux density of 0.5 T? What are the flux densities of the four sides of the core at that current? What is the total flux present in the air gap? 12 SOLUTION The magnetization curve for this core is shown below: An air-gap flux density of 0.5 T requires a total flux of      eff 0.5 T 0.05 m 0.05 m 1.05 0.00131 Wb   BA   This flux requires a flux density in the right-hand leg of 13    right 0.00131 Wb 0.524 T 0.05 m 0.05 m B A     The flux density in the other three legs of the core is    top left bottom 0.00131 Wb 0.262 T 0.10 m 0.05 m B B B A       The magnetizing intensity required to produce a flux density of 0.5 T in the air gap can be found from the equation ag ag o B   H : ag ag 7 0 0.5 T 398 kA t/m 4 10 H/m B H         The magnetizing intensity required to produce a flux density of 0.524 T in the right-hand leg of the core can be found from Figure P1-9 to be right 410 A t/m H   The magnetizing intensity required to produce a flux density of 0.262 T in the top, left, and bottom legs of the core can be found from Figure P1-9 to be top left bottom 240 A t/m H H H     The total MMF required to produce the flux is TOT ag ag right right top top left left bottom bottom H l H l H l H l H l      F          TOT 398 kA t/m 0.0005 m 410 A t/m 0.40 m 3 240 A t/m 0.40 m       F TOT 278.6 164 288 651 A t     F  and the required current is TOT 651 A t 0.651 A 1000 t i N     F The flux densities in the four sides of the core and the total flux present in the air gap were calculated above. 1-15. A transformer core with an effective mean path length of 6 in has a 200-turn coil wrapped around one leg. Its cross-sectional area is 0.25 in2, and its magnetization curve is shown in Figure 1-10c. If current of 0.3 A is flowing in the coil, what is the total flux in the core? What is the flux density? 14 SOLUTION The magnetizing intensity applied to this core is       200 t 0.3 A 394 A t/m 6 in 0.0254 m/in c c Ni H l l      F From the magnetization curve, the flux density in the core is B 1.35 T The total flux in the core is    2 2 0.0254 m 1.35 T 0.25 in 0.000218 Wb 1 in BA           1-16. The core shown in Figure P1-2 has the flux  shown in Figure P1-12. Sketch the voltage present at the terminals of the coil. 15 SOLUTION By Lenz’ Law, an increasing flux in the direction shown on the core will produce a voltage that tends to oppose the increase. This voltage will be the same polarity as the direction shown on the core, so it will be positive. The induced voltage in the core is given by the equation ind d e N dt   so the voltage in the windings will be Time dt N d  eind 0 < t < 2 s  500 t 0.010 Wb 2 s 2.50 V 2 < t < 5 s  500 t 0.020 Wb 3 s  -3.33 V 5 < t < 7 s  500 t 0.010 Wb 2 s 2.50 V 7 < t < 8 s  500 t 0.010 Wb 1 s 5.00 V The resulting voltage is plotted below: 16 1-17. Figure P1-13 shows the core of a simple dc motor. The magnetization curve for the metal in this core is given by Figure 1-10c and d. Assume that the cross-sectional area of each air gap is 18 cm2 and that the width of each air gap is 0.05 cm. The effective diameter of the rotor core is 5 cm. (a) We wish to build a machine with as great a flux density as possible while avoiding excessive saturation in the core. What would be a reasonable maximum flux density for this core? (b) What would be the total flux in the core at the flux density of part (a)? (c) The maximum possible field current for this machine is 1 A. Select a reasonable number of turns of wire to provide the desired flux density while not exceeding the maximum available current. 17 SOLUTION The magnetization curve for this core is shown below: The relative permeability of this core is shown below: Note: This is a design problem, and the answer presented here is not unique. Other values could be selected for the flux density in part (a), and other numbers of turns could be selected in part (c). These other answers are also correct if the proper steps were followed, and if the choices were reasonable. (a) From Figure 1-10c, a reasonable maximum flux density would be about 1.2 T. Notice that the saturation effects become significant for higher flux densities. (b) At a flux density of 1.2 T, the total flux in the core would be (1.2 T)(0.05 m)(0.05 m) 0.0030 Wb   BA   (c) The total reluctance of the core is: 18 TOT stator air gap 1 rotor air gap 2     R R R R R At a flux density of 1.2 T, the relative permeability r  of the stator is about 3800, so the stator reluctance is      stator stator 7 stator stator 0.60 m 50.3 kA t/Wb 3800 4 10 H/m 0.05 m 0.05 m l A         R At a flux density of 1.2 T, the relative permeability r  of the rotor is 3800, so the rotor reluctance is      rotor rotor 7 stator rotor 0.05 m 4.2 kA t/Wb 3800 4 10 H/m 0.05 m 0.05 m l A         R The reluctance of both air gap 1 and air gap 2 is    air gap air gap 1 air gap 2 7 2 air gap air gap 0.0005 m 221 kA t/Wb 4 10 H/m 0.0018 m l A          R R Therefore, the total reluctance of the core is TOT stator air gap 1 rotor air gap 2     R R R R R TOT 50.3 221 4.2 221 496 kA t/Wb       R The required MMF is    TOT TOT 0.003 Wb 496 kA t/Wb 1488 A t     F R  Since F  Ni , and the current is limited to 1 A, one possible choice for the number of turns is N = 2000. This would allow the desired flux density to be achieved with a current of about 0.74 A. 1-18. Assume that the voltage applied to a load is 208 30 V     V and the current flowing through the load is 2 20 A .    I (a) Calculate the complex power S consumed by this load. (b) Is this load inductive or capacitive? (c) Calculate the power factor of this load? (d) Calculate the reactive power consumed or supplied by this load. Does the load consume reactive power from the source or supply it to the source? SOLUTION (a) The complex power S consumed by this load is      * 208 30 V 2 20 A 208 30 V 2 20 A               * S VI  416 50 VA     S (b) This is a capacitive load. (c) The power factor of this load is   PF cos 50 0.643 leading     (d) This load supplies reactive power to the source. The reactive power of the load is      sin 208 V 2 A sin 50 319 var Q VI        1-19. Figure P1-14 shows a simple single-phase ac power system with three loads. The voltage source is  240 0 V , impedances of these three loads are   V 1 10 30    Z 2 10 45    Z 3 10 90      Z 19 Answer the following questions about this power system. (a) Assume that the switch shown in the figure is initially open, and calculate the current I, the power factor, and the real, reactive, and apparent power being supplied by the source. (b) How much real, reactive, and apparent power is being consumed by each load with the switch open? (c) Assume that the switch shown in the figure is now closed, and calculate the current I, the power factor, and the real, reactive, and apparent power being supplied by the source. (d) How much real, reactive, and apparent power is being consumed by each load with the switch closed? (e) What happened to the current flowing from the source when the switch closed? Why? + - I V + - + - + - Z1 Z2 3 Z  120 0 V   V SOLUTION (a) With the switch open, only loads 1 and 2 are connected to the source. The current in Load 1 is 1I 1 240 0 V 24 30 A 10 30 A         I  The current in Load 2 is 2I 2 240 0 V 24 45 A 10 45 A         I  Therefore the total current from the source is 1 2 24 30 A 24 45 A 47.59 37.5 A               I I I The power factor supplied by the source is   PF cos cos 37.5 0.793 lagging      Note that the angle  used in the power factor and power calculations is the impedance angle, which is the negative of the current angle as long as voltage is at 0°. The real, reactive, and apparent power supplied by the source are      cos 240 V 47.59 A cos 37.5 9061 W P VI           cos 240 V 47.59 A sin 37.5 6953 var Q VI         240 V 47.59 A 11,420 VA S VI   (b) The real, reactive, and apparent power consumed by Load 1 are      cos 240 V 24 A cos 30 4988 W P VI           cos 240 V 24 A sin 30 2880 var Q VI         cos 240 V 24 A 5760 VA S VI     The real, reactive, and apparent power consumed by Load 2 are      cos 240 V 24 A cos 45 4073 W P VI      20      cos 240 V 24 A sin 45 4073 var Q VI         cos 240 V 24 A 5760 VA S VI     As expected, the real and reactive power supplied by the source are equal to the sum of the real and reactive powers consumed by the loads. (c) With the switch closed, all three loads are connected to the source. The current in Loads 1 and 2 is the same as before. The current in Load 3 is 3I 3 240 0 V 24 90 A 10 90 A         I  Therefore the total current from the source is 1 2 3 24 30 A 24 45 A 24 90 A 38.08 7.5 A                   I I I I The power factor supplied by the source is   PF cos cos 7.5 0.991 lagging      The real, reactive, and apparent power supplied by the source are      cos 240 V 38.08 A cos 7.5 9061 W P VI           cos 240 V 38.08 A sin 7.5 1193 var Q VI         240 V 38.08 A 9140 VA S VI   (d) The real, reactive, and apparent power consumed by Load 1 are      cos 240 V 24 A cos 30 4988 W P VI           cos 240 V 24 A sin 30 2880 var Q VI         cos 240 V 24 A 5760 VA S VI     The real, reactive, and apparent power consumed by Load 2 are      cos 240 V 24 A cos 45 4073 W P VI           cos 240 V 24 A sin 45 4073 var Q VI         cos 240 V 24 A 5760 VA S VI     The real, reactive, and apparent power consumed by Load 3 are      cos 240 V 24 A cos 90 0 W P VI            cos 240 V 24 A sin 90 5760 var Q VI           cos 240 V 24 A 5760 VA S VI     As expected, the real and reactive power supplied by the source are equal to the sum of the real and reactive powers consumed by the loads. (e) The current flowing decreased when the switch closed, because most of the reactive power being consumed by Loads 1 and 2 is being supplied by Load 3. Since less reactive power has to be supplied by the source, the total current flow decreases. 1-20. Demonstrate that Equation (1-59) can be derived from Equation (1-58) using simple trigonometric identities:   ( ) ( ) ( ) 2 cos cos p t v t i t VI t t       (1-58)   ( ) cos 1 cos2 sin sin2 p t VI t VI t        (1-59) 21 SOLUTION The first step is to apply the following identity:    1 cos cos cos cos 2                The result is   ( ) ( ) ( ) 2 cos cos p t v t i t VI t t       )     1 ( ) 2 cos cos 2 p t VI t t t t                         ( ) cos cos 2 p